Question

Rút gọn:

\(A=\frac{\sqrt{1-\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}\right]}{2-\sqrt{1-x^2}}\)

Answer 1

Lời giải:

Đặt \(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} a^2b^2=1-x^2\\ a^2+b^2=2\end{matrix}\right.\)

\(\Rightarrow A=\frac{\sqrt{1-ab}(a^3+b^3)}{2-ab}\)

\(=\frac{\sqrt{1-ab}(a+b)(a^2-ab+b^2)}{2-ab}\)

\(=\frac{\sqrt{1-ab}(a+b)(2-ab)}{2-ab}=\sqrt{1-ab}(a+b)\)

\(=\sqrt{\frac{2-2ab}{2}}(a+b)=\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)\)

\(=\frac{|a-b|}{\sqrt{2}}(a+b)=\frac{|a^2-b^2|}{\sqrt{2}}=\frac{|(1+x)-(1-x)|}{\sqrt{2}}=\sqrt{2}|x|\)